Question:

The radius of a sphere is measured to be (7.50 ± 0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest integer, is __________.

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The relative error in a calculated quantity is the sum of relative errors of components multiplied by their respective powers.
Updated On: Jan 12, 2026
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Correct Answer: 34

Solution and Explanation

Step 1: Volume of a sphere \(V = \frac{4}{3} \pi r^3\).
Step 2: Calculate the percentage error in volume (\(\frac{\Delta V}{V} \times 100\)). For a power relation \(V \propto r^3\), the relative error is: \[ \frac{\Delta V}{V} = 3 \frac{\Delta r}{r} \]
Step 3: Substitute the given values. \(\Delta r = 0.85\), \(r = 7.50\). \[ % \text{ Error} = 3 \times \left( \frac{0.85}{7.50} \right) \times 100 \] \[ % \text{ Error} = 3 \times (0.1133) \times 100 = 34% \]
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